Solubility Product Constant Calculation

Question

The dissolution of solid calcium fluoride is represented by the equilibrium equation below.
$$	ext{CaF}_2 (s) ightleftharpoons 	ext{Ca}^{2+}(aq) + 2	ext{F}^{-}(aq)$$
Given the equilibrium concentrations in the table, what is the value of the solubility product constant, $$K_{sp}$$, for $$	ext{CaF}_2$$?

| Ion                 | Equilibrium Concentration (M) |
|---------------------|-------------------------------|
| $$	ext{Ca}^{2+}$$  | 1.0e-4                        |
| $$	ext{F}^{-}$$    | 2.0e-4                        |

Accepted Answer

$$4.0 	imes 10^{-12}$$

Answer

$$8.0 	imes 10^{-12}$$

Answer

$$2.0 	imes 10^{-8}$$

Answer

$$1.0 	imes 10^{-4}$$

Rank	User	Correct Count	Attempt Count	Total Time	Score
#1	gwantsai	2	2	0m 00s	200
#2	itzzhanachan	1	1	0m 00s	100
#3	zainabbkhokharr	1	1	0m 00s	100
#4	kkat.s.music	1	1	0m 37s	63
#5	Pelisx	1	2	1m 26s	4
#6	ireneandrea449	0	1	1m 56s	-126
#7	Derinchiay2013	4	6	59m 06s	-3,166

Solubility Product Constant Calculation

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